Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 8}{x + 7} = \dfrac{6x + 24}{x + 7}$
Solution: Multiply both sides by $x + 7$ $ \dfrac{x^2 + 8}{x + 7} (x + 7) = \dfrac{6x + 24}{x + 7} (x + 7)$ $ x^2 + 8 = 6x + 24$ Subtract $6x + 24$ from both sides: $ x^2 + 8 - (6x + 24) = 6x + 24 - (6x + 24)$ $ x^2 + 8 - 6x - 24 = 0$ $ x^2 - 16 - 6x = 0$ Factor the expression: $ (x - 8)(x + 2) = 0$ Therefore $x = 8$ or $x = -2$ The original expression is defined at $x = 8$ and $x = -2$, so there are no extraneous solutions.